多个设置超时同时运行

Question

所以，我有一个for循环，我想一次运行多个设置超时，所以它可以不断地检查它是否是一个特定的时间多次。我不能一次检查所有的多次，检查需要在不同的时间后重复。它只继续重复循环的最后一次迭代，即lat中的最后一个字典。

我的方法简化了一点:

lat = list of dictionaries with some values that differentiate them + times
for(i = 0; i< lat.length; i++){
if(lat[i][differentiate value]) = {
function checktime(){
if(currenttime != lat[i]){
setTimeout(checktime,30000)
}
else{
console.log("yay the time is right!")
}
}
else if(lat[i][differentiate value]) = {
function checktime(){
if(currenttime != lat[i]){
setTimeout(checktime,50000)
}
else{
console.log("yay the time is right!")
}
}
}

我该怎么做呢(这是一个通知应用程序)

原始代码:(每个值在后面看起来像[[75年,null,[2021] 7日,28日],null,[[9 52"p"],"e& r")的通知,预定日期,这些提醒的日期,时间,这将是第一个,如果和数组。为了测试，我有2个不同时间的later):

chrome.storage.sync.get('later', function(data){
if (data.later !== null){
var lat = data.later;

for(i = 0; i< lat.length; i++){
var currentDict = lat[i];

if(currentDict['notis'][1] !== null){
console.log("cheese")
var by = currentDict['notis'][1];
console.log(by)
const d = new Date();
var hr = d.getHours();
var min = d.getMinutes();
var sec = d.getSeconds();

var da = currentDict['notis'][0][2][1];
var mo = currentDict['notis'][0][2][0];
var ye = currentDict['notis'][0][2][2]
var h = by[0];
var m = by[1];
var ampm = by[2];
if(ampm == "p"){
h = h + 12;
}
var byMS = h*3600000 + m*60000;
var currentMS = hr*3600000 + min*60000 + sec*1000;
//check if right date then check if time is lesss than

function checkdate(){
var day = d.getDate();
var month = d.getMonth() + 1;
var year = d.getFullYear();
if(da == day && mo == month && ye == year){
var amt = 0;
function checktime(){
if(byMS >= currentMS){
//noti and delete
var int = setInterval(function(){
chrome.notifications.create({
title: currentDict['name'],
message: "do da " + currentDict['name'],
iconUrl: "logo.png",
type: "basic"

})

amt++
console.log(amt)
const dddd = new Date();
console.log(dddd.getMinutes() + " / " + dddd.getSeconds())
if(amt >= currentDict['notis'][0][0]){
clearInterval(int)
console.log("done")
//ju finish taht
console.log(lat)
lat.splice(lat.indexOf(currentDict),1)
console.log(lat)
chrome.storage.sync.set({'later': lat})

}
}, (byMS-currentMS)/currentDict['notis'][0][0])
}
else{
setTimeout(checktime,30000)
}
}
checktime();
}
else{
setTimeout(checkdate,66400000)
}
}
checkdate();
}
else if(currentDict['notis'][2] !== null){

console.log("cheese")
var at = currentDict['notis'][2];
console.log(at)

var arrayat = [];
for(j = 0; j<= at.length-1; j++){
var atcycle = at[j];
console.log(atcycle)
const ddd = new Date();
var hr = ddd.getHours();
var min = ddd.getMinutes();

var da = currentDict['notis'][0][2][1];
var mo = currentDict['notis'][0][2][0];
var ye = currentDict['notis'][0][2][2]
var hrat = atcycle[0];
var minat = atcycle[1];
var ampm = atcycle[2];
if(ampm == "p"){
hrat = hrat + 12;
}
console.log(hrat + "/" + minat + "/" + ampm)
if(hr <= hrat){

var temparray = [];
temparray.push(hrat)
temparray.push(minat)
arrayat.push(temparray);
console.log(arrayat)
}

else if(hr == hrat){
if(min<minat){
var temparray = [];
temparray.push(hrat)
temparray.push(minat)
arrayat.push(temparray);
console.log(arrayat)}
}       
}      
console.log(arrayat.length)     
function checkdate(){
console.log(arrayat.length)
console.log("date")
const d = new Date();
var day = d.getDate();
var month = d.getMonth() + 1;
var year = d.getFullYear();
if(da == day && mo == month && ye == year){
function check(){
console.log(arrayat.length)
console.log("check")
for(l=0; l<arrayat.length; l++){
console.log(arrayat.length)
const dd = new Date();
var hr = dd.getHours();
var min = dd.getMinutes();
console.log(arrayat[l][1])
console.log(min)
if(arrayat[l][0] == hr && arrayat[l][1] == min ){ //at one of the times
console.log(arrayat)
arrayat.splice(l,1)
console.log(arrayat)
if(arrayat.length == 0){
lat.splice(lat.indexOf(currentDict),1)
chrome.storage.sync.set({'later': lat})
console.log(lat)

}
chrome.notifications.create({
title: currentDict['name'],
message: "do da " + currentDict['name'],
iconUrl: "logo.png",
type: "basic"

})
//add noti with name and delete it
console.log(arrayat)
check();
}
}
console.log(arrayat.length)
if(arrayat.length !== 0){
console.log("and repeat")
setTimeout(check,15000);//SETINTERVAL INSTEAD? ANDCLEAR
}

}
check();
}
else{
setTimeout(checkdate,66400000)
}
}
checkdate();

}

}
}

})
}

Answer 1

这是错误的方法。你知道时间，所以你知道它们在时间上有多远。如果时间到了，你不需要一遍又一遍地检查。相反，计算出距离所讨论的时间有多少毫秒，然后为该时间设置一个超时。

Answer 2

您基本上需要将不同的函数传递给每个setTimeout():

function checktime1(){
if(currenttime != lat[i]){
setTimeout(checktime1,30000)
}
else{
console.log("yay the time is right!")
}
}
function checktime2(){
if(currenttime != lat[i]){
setTimeout(checktime2,50000)
}
else{
console.log("yay the time is right!")
} 
}

然而，这显然是不可伸缩的。如果你有很多超时，你就会写很多重复的函数。更糟的是，如果您有一个动态的超时次数，这将无法解决您的问题。

幸运的是，javascript允许我们在另一个函数中声明函数:

function checktime(timeout){
function loop () {
if(currenttime != lat[i]){
setTimeout(loop,timeout)
}
else{
console.log("yay the time is right!")
}
}
loop();
}

这允许您根据需要动态创建函数:

checktime(30000);
checktime(50000);

如果您需要向每个超时传递参数(例如;如果你需要显示自定义警报)，你可以把它传递给你的外部函数，它将在闭包中被捕获:

function checktime(timeout, message){
function loop () {
if(currenttime != lat[i]){
setTimeout(loop,timeout)
}
else{
console.log(message)
}
}
loop();
}

那么你可以这样做:

checktime(30000, "Yay, the time is right for timer 1");
checktime(50000, "Yay, the time is right for timer 2");

同样，如果你需要为每个计时器执行自定义逻辑，你也可以将其作为参数(回调)传递:

function checktime(timeout, callback){
function loop () {
if(currenttime != lat[i]){
setTimeout(loop,timeout);
}
else{
console.log("yay the time is right!");
callback(); // <----------------- execute custom logic
}
}
loop();
}

可以这样使用:

checktime(30000, () -> console.log("TIMER 1"));
checktime(50000, () -> alert("TIMER 2"));